Question

Write a quadratic polynomial whose zrroes are √3+1 ,√3-1

Answer 1

Solution

Given :-

• Zeroes of quadratic polynomial is ( √3 + 1) & (√3 - 1)

Find :-

• Equation of quadratic polynomial

Explanation,

Let,

• Zeroes of quadratic polynomial be p & q

So,

==> Sum of zeroes = (√3+1) + (√3-1)

==> Sum of zeroes = 2√3

==> ( p + q) = 2√3

Again,

==> product of zeroes = ( √3+1) × (√3-1)

==> product of zeroes = (√3)² - 1²

==> product of zeroes = (3-1)

==> p . q = 2

Formula of Equation,

★ x² - ( p + q)x + (p . q) = 0

Keep above values

==> x² - (2√3)x + 2 = 9

Hence, Required Formula

• x² - 2√3x + 2 = 0

_________________

Answer 2

$\sf\large{\underline{\purple{\underline{\red{Question:-}}}}}$

Write a quadratic polynomial whose zrroes are √3+1 ,√3-1

$\sf\large{\underline{\purple{\underline{\red{we \:know:-}}}}}$

$\sf\large{\underline{\purple{\underline{\red{General\: equation:-}}}}}$

$\sf→{\fbox{\purple{\underline{\red{x^2-(\alpha+\beta)x+\alpha\beta=0}}}}}$

Let the zeroes be = $\alpha/:and\:\beta$

$\sf→sum\: of \:zeroes \alpha+\beta\\\sf→ product\:of\:zeroes=\alpha\beta$

$\sf\large{\underline{\purple{\underline{\red{So,:-}}}}}$

$\sf→sum \:of \:zeroes =( \sqrt3+1)+(\sqrt3+1)\\\sf→ 2\sqrt3$

$\sf\large{\underline{\purple{\underline{\red{And:-}}}}}$

$\sf→Product\: of\: zeroes=( \sqrt3+1)×(\sqrt3+1)\\\sf→ 2$

$\sf\large{\underline{\purple{\underline{\red{Hence:-}}}}}$

$\sf\large{\underline{\purple{\underline{\red{By\:General\: equation:-}}}}}$

$\sf→ x^2-(2\sqrt3)x+2=0\\\sf→ x^2-2\sqrt3x×2=0$

$\large\sf→{\fbox{\red{\underline{x^2-2\sqrt3x+2}}}}$